Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. Calculate the quantity of alcohol in a 240 mL bottle of the oral solution. (Round to two decimal places.)

Answer: 0.35 moles

Explanation:

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]   (At constant temperature and pressure)  

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 52.0 L

[tex]V_2[/tex] = final volume of gas = 13.0 L

[tex]n_1[/tex] = initial number of moles = 1.40

[tex]n_2[/tex] = final number of moles = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{52.0}{1.40}=\frac{13.0}{n_2}[/tex]

[tex]n_2=0.35moles[/tex]

Therefore, the moles of the gas that will remain is 0.35 moles.

The acceptable IUPAC name among the given options is 3,6-dimethylheptane, as it correctly follows the IUPAC rules for nomenclature where similar substituents are denoted with prefixes like 'di-', 'tri-' etc.

The correct answer is option (A) 3,6-dimethylheptane. The IUPAC rules for nomenclature of organic compounds mandate that when there are multiple substituents of the same type, as in this case where we have two methyl groups, the carbon atoms to which they are attached are indicated by commas between the numbers and preceded by prefixes such as 'di-', 'tri-' etc. Thus, 3,6-dimethylheptane follows these rules, making it an acceptable IUPAC name. The other options do not follow the IUPAC rules for nomenclature.

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According to IUPAC nomenclature rules for organic compounds, option (B) 2-ethyl-3-methylheptane is the correct name. This indicates a heptane chain with an ethyl group on the second carbon and a methyl group on the third carbon.

The question pertains to the correct naming of organic compounds according to the system laid out by the International Union of Pure and Applied Chemistry (IUPAC). According to IUPAC nomenclature, the longest carbon chain in the molecule is identified first, and is named using Greek prefixes indicating the number of carbon atoms. All substituents (any atoms or groups of atoms) that are attached to the longest chain are then identified and designated with prefixes and the position on the carbon chain.

After examining the given options, Option (B) 2-ethyl-3-methylheptane is the correct IUPAC name. This indicates that the longest carbon chain in the molecule is heptane (7 carbons), with an ethyl group (2 carbons) attached to the second carbon and a methyl group (1 carbon) attached to the third carbon. The other choices (3,6-dimethylheptane; 5-methyl-3-ethylheptane; and 4-ethyl-4-methylheptane) do not follow the IUPAC rules of nomenclature appropriately, either because they involve numbering from the wrong end of the carbon chain or because identical groups on the same carbon are not correctly identified.

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Answer: [tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.

A general single displacement reaction can be represented as :

[tex]A+BC\rightarrow AC+B[/tex]

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced equation for the reaction of aluminum with copper(II) sulfate solution is:

[tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]

The balanced equation for the reaction between aluminum and copper(II) sulfate solution is: 2Al (s) + 3CuSO4 (aq) -> Al2(SO4)3 (aq) + 3Cu (s). In this reaction, aluminum displaces copper to form aluminum sulfate, and copper is precipitated out.

The question asks for the balanced equation for the reaction of aluminum with copper(II) sulfate solution. When aluminum reacts with copper(II) sulfate solution, it displaces copper to form aluminum sulfate, and copper is precipitated out.

The unbalanced equation for the reaction is:
Al (s) + CuSO4 (aq) -> Al2(SO4)3 (aq) + Cu (s)

To balance it, add stoichiometric coefficients to the reactants and products:'

2Al (s) + 3CuSO4 (aq) -> Al2(SO4)3 (aq) + 3Cu (s)

This means that 2 moles of aluminum react with 3 moles of copper(II) sulfate to produce 1 mole of aluminum sulfate and 3 moles of copper.

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Answer:

Composition of the mixture:

[tex]x_{A} =0.652=65.2[/tex] %

[tex]x_{B} =0.348=34.8[/tex] %

Composition of the vapor mixture:

[tex]y_{A} =0.809=80.9[/tex]%

[tex]y_{B} =0.191=19.1[/tex]%

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with [tex]x_{A}[/tex] and [tex]x_{B}[/tex] molar fractions can be calculated as:

[tex]P_{vap}=x_{A}P_{A}+x_{B}P_{B}[/tex]

Where [tex]P_{A}[/tex] and [tex]P_{B}[/tex] are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when [tex]P_{vap}=P[/tex]. When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

[tex]0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652[/tex]

With the same assumptions, the vapor mixture may obey to the equation:

[tex]x_{A}P_{A}=y_{A}P[/tex], where P is the total pressure and y is the fraction in the vapor phase, so:

[tex]y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9[/tex] %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

Answer:

Must be either ether or ester.

Explanation:

The incorrect statement is that the monosaccharides must ether or esters.

The main and compulsory functional group in a monosaccharide are aldehyde and ketones.

The other statements are true

1) smallest monosaccharides are composed of three carbons: the smallest monosaccharide is glyceraldehyde.

a) aldoses and ketoses : they have either aldehyde group or ketone group

example: glucose is aldose and fructose if keotse

c) exist in many isomeric forms: there can be many arrangement on chiral carbons thus many stereoisomers are possible

e) Two monosaccharides can combine to form a disaccharide: yes for example glucose combines with fructose to give sucrose.

Answer:

a) [tex]1.065\times 10^{-8} [/tex]fraction of carbon−14 in a piece of charcoal remains after 14.0 years.

b) [tex]0.000\times 10^{-3} [/tex]fraction of carbon−14 in a piece of charcoal remains after  [tex]1.900\times 10^4 years[/tex]

c) [tex]0.0000\times 10^{-4} [/tex]fraction of carbon−14 in a piece of charcoal remains after  [tex]1.0000\times 10^5 years[/tex].

Explanation:

The fraction of a radioactive isotope remaining at time t is given by:

[tex][A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}[/tex]

Taking log both sides:

[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]

[A] =  fraction at given time t

[tex]t_{\frac{1}{2}}[/tex] = half life of the carbon−14 =5,730 years

a)When , t = 14 years

[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]

[tex]\log [A]= 14 years\times (-3010)-\log [5,730 years][/tex]

[tex][A]=1.065\times 10^{-8} [/tex]

b)When , t = [tex]1.900\times 10^4 years[/tex]

[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]

[tex]\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years][/tex]

[tex][A]=0.000\times 10^{-3} [/tex [/tex]

c)When , t = [tex]1.0000\times 10^5 years[/tex]

[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]

[tex]\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years][/tex]

[tex][A]=0.0000\times 10^{-4} [/tex]

Answer: 0.11 M

Explanation:

The equation for the reaction is given as:

[tex]2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O[/tex]

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = ?

[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = 0.105 L= 105 ml

[tex]M_2[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution = 0.2 M

[tex]V_2[/tex] = volume of [tex]Ba(OH)_2[/tex] solution = 29.1 ml

[tex]n_1[/tex] = valency of [tex]HNO_3[/tex] = 1

[tex]n_2[/tex] = valency of [tex]Ba(OH)_2[/tex] = 2

[tex]1\times M_1\times 105=2\times 0.2\times 29.1[/tex]

[tex]M_1=0.11M[/tex]

Therefore, the concentration of [tex]HNO_3[/tex] will be 0.11 M.

Answer:

PV=nRT

n = PV/RT

n = m/Mm

m/Mm = PV/RT

m = MmPV/RT

T in kelvin = T Celsius + 273.15 = 293.15 K

m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)

mass in grams = 82.8 grams  

Explanation:

Ideal gases formula is PV=nRT, where:

P is the pressure (1.39 atm in this case)

V is the volume (55.0 L in this case)

R is the gas constant (0.08206 L.atm/K.mole)

T is the temperature (20.0C) should be converted to Kelvin

all the unit should correspond to the one in the R.

we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.

we found the mass to be 82.8 grams

PV=nRT is the formula for ideal gases, where P is the pressure and R is the constant and T is the temperature (1.39 atm in this case)

[tex]m = MmPV/RT\\m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)[/tex]

Thus, the mass was discovered to be 82.8 grams.

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Answer:

At 1692.31 K.

Explanation:

The Gibbs free energy is a measurement of the energy available for the system to realize a change. And can be calculated by the relation of enthalpy, temperature and entropy by the Gibbs equation:

[tex]\Delta G =  \Delta H + T\Delta S[/tex]

A procces will be spontaneous if the Gibbs free energy of the system is negative.

So, if we set the ΔG of the system as -.1 J to make it so that the proccess will be spontanteous and input the given values of entropy and enthalpy into the Gibbs equation, we get:

[tex]-.01 J = 22,000 J - T(-13 J/K)[/tex]

[tex]T = 1692.3 K[/tex]

Explanation:

We have given the acid that is, acetic acid. It is known that acetic acid is a weak acid.

Therefore, formula that depicts relationship between pH, [tex]pK_{a}[/tex] and weak acid is as follows.

                  pH = [tex]pK_{a}[/tex] + log[tex]\frac{[A^{-}]}{[HA]}[/tex]

where,          [HA] = concentration of weak acid

      [tex][A^{-}][/tex] = concentration of conjugate base of given weak acid

Since, we have to calculate the concentration of conjugate base of acetic acid. So, let it be equal to x. Whereas concentration of acetic acid is 10 mmol, [tex]pK_{a}[/tex] is 4.74 and pH is 5.03.

Hence, putting these values in the above formula as follows.

                  pH = [tex]pK_{a}[/tex] + log[tex]\frac{[A^{-}]}{[HA]}[/tex]

                  5.03 = 4.74 + log[tex]\frac{x}{10}[/tex]

              [tex]\frac{x}{10}[/tex] = antilog (0.29)

                      x = 19.4 mmol

Thus, we can conclude that there is 19.4 mmol acetate (the conjugate base of acetic acid) will be needed to add to this given solution.

To produce a buffer solution with a pH of 5.03 using acetic acid, you need to calculate the concentration of acetic acid and acetate ion. By applying the Henderson-Hasselbalch equation, you can determine the concentration of acetate needed in the buffer solution.

To produce a buffer solution with a pH of 5.03, you already have a solution containing 10 mmol of acetic acid. The first step is to calculate the concentration of acetic acid in the initial solution. Using the formula [H3O+] = √(√Ka x [CH3CO2H]), we find that [H3O+] = √(√1.8 x 10^-5 x 0.01) = 0.00113 M.

Next, we need to determine the concentration of acetate required. Since pH changes by 1 unit when the acetic acid concentration is reduced to 11% of the acetate ion concentration, we can calculate the concentration of acetate using the equation 0.11 = (0.00113-x)/(0.01+x), where x is the concentration of acetate.

Solving this equation, we can find x, which represents the concentration of acetate needed in the buffer solution.

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Explanation:

For transfer of heat between any two substances or matter can be occurred by conduction or convection process.

Conduction is a process in which two objects of different temperature are placed in contact with each other. Therefore, heat transfers from hotter object to colder object until the temperature of both objects becomes equal.

For example, a metal spoon placed in a hot cup of tea.

Convection is a process in which a fluid generally liquid or gas is heated and denser material (colder) sinks at the bottom whereas less denser material (hotter) rises at the top. This causes conventional currents in the fluid.

For example, heating rice in a pot full of water.

Answer: The mass of oxygen gas required will be 6.648 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     ....(1)

Given mass of hydrochloric acid = 30.3 g

Molar mass of hydrochloric acid = 36.46 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }HCl=\frac{30.3g}{36.46g/mol}=0.831mol[/tex]

For the given chemical reaction, the balanced equation follows:

[tex]4HCl+O_2\rightarrow 2H_2O+2Cl_2[/tex]

By Stoichiometry of the reaction:

4 moles of HCl reacts with 1 mole of oxygen gas.

So, 0.831 moles of HCl will react with = [tex]\frac{1}{4}\times 8.31=0.20775mol[/tex] of oxygen gas.

Now, calculating the mass of oxygen gas by using equation 1, we get:

Moles of oxygen gas = 0.20775 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

[tex]0.20775mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=6.648g[/tex]

Hence, the mass of oxygen gas required will be 6.648 grams.

Answer:

Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.

hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond  must be oriented in  the opposite directions with respect to each other.]

So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.

If one isomer of the alkene is trans then the other two isomers may be cis .

Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.

The two possibility of cis structures are possible:

in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.

Or the other way could be that two  chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.

Kindly refer the attachments for the structure of compounds:

Compound X is a trans-form of dichloroethylene (ClCH=CHCl), compound Z, which has a dipole moment, is a cis-form of dichloroethylene (ClC=CH2) and, compound Y is ClCH=CH2.

The structures of the three different dichloroethylenes designated as X, Y, and Z can be understood based on their dipole moments and their reactivity with hydrogen. Compound X, which has no dipole moment, must have a symmetrical structure with the chlorines on opposite sides of the double bond, resulting in a trans conformer. This could be represented as ClCH=CHCl. Since Compounds X and Z each combine with hydrogen to give the same product (ClCH2―CH2Cl), Z must also possess this structure; however, it has a dipole moment. Therefore, Z has the chlorines on the same side of the double bond, resulting in a cis conformer. This structure can be represented as ClC=CH2. The remaining isomer, Y, must therefore have one chlorine atom at each carbon, represented as ClCH=CH2.

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Answer: Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

The chemical equation for the combustion of ethylene in nitrous oxide follows the equation:

[tex]C_2H_4+N_2O\rightarrow 2CO_2+2H_2O+6N_2[/tex]

By stoichiometry of the reaction;

6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.

Calculating the mass of nitrous oxide using equation 1:

Molar mass of nitrous oxide = 44 g/mol

Moles of nitrous oxide = 6 moles

Putting values in equation 1, we get:

[tex]6mol=\frac{\text{Mass of nitrous oxide}}{44g/mol}\\\\\text{Mass of nitrous oxide}=264g[/tex]

Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.

The chemical equation for the combustion of ethylene in air follows the equation:

[tex]C_2H_4+3O_2\rightarrow 2CO_2+2H_2O[/tex]

By stoichiometry of the reaction;

3 moles of oxygen are required for the complete combustion of ethylene molecule.

Calculating the mass of oxygen using equation 1:

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 3 moles

Putting values in equation 1, we get:

[tex]3mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=96g[/tex]

Thus, 96 grams of oxygen are required for the complete combustion of ethylene.

Answer:

A: 1,2-dimethylcyclopropane

Explanation:

The possible cyclic structure with formula C₅H₁₀ are shown in the image.

A is a cyclic compound. On monochlorination, A yields 3 products.

To have 3 products on monochlorination, there should be three different carbon atoms.

Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.

Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.

Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.

Considering structure 5, there are three different carbon atoms,  thus three different structure are formed and thus the A is structure 5.

The monochlorination products are 1,2-dimethylcyclopropane, 1,3-dichlorocyclopentane, and  1,1-dichlorocyclopentane.

Monochlorination is the process in which substitution of one atom with chlorine atoms occur.

Given,

When [tex]C_5H_1_0[/tex] undergoes chlorination, it yields three different constitutional isomers.

So, the possible molecules are pentene and cyclopentane.

By monochlorination we get only one product that is 1-chlorocyclepentane.

By dichlorination of the compound, we get three products that are 1,2 -dichlorocyclopentane, 1,3-dichlorocyclopentane, and  1,1-dichlorocyclopentane respectively as given in the diagram.

Thus, the products are 1,2 -dichlorocyclopentane, 1,3-dichlorocyclopentane, and  1,1-dichlorocyclopentane.

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Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

The reaction between sodium tert-butoxide and 2-chlorobutane is expected to yield 2-butene and 1-butene as the major organic products, with 2-butene being more predominant due to Hofmann Elimination with the bulky base.

When sodium tert-butoxide (NaOC(CH₃)₃) reacts with 2-chlorobutane, it acts as a bulky Bronsted-Lowry base. Due to its steric bulk, it tends to remove a hydrogen from the less hindered (more accessible) carbon adjacent to the carbon bearing the chlorine, which in this case is the carbon at the 3-position. As a result, the major organic product(s) of this reaction are expected to be alkenes, specifically 2-butene and, to a lesser extent, 1-butene due to the Hofmann Elimination favoring the less substituted alkene when considering a bulky base like sodium tert-butoxide.

Answer:The frequency of absorption for the proton having chemical shift 1.48 ppm is 740 Hz downfield from TMS.

The frequency of absorption for the proton having chemical shift 3.57 ppm is 1785 Hz downfield from TMS.

Explanation:

We are given with the following data:

Frequency of the instrument(NMR spectrometer)=500MHz=500×10⁶Hz

Chemical shift (δ) value  for 1st proton=1.48PPM=1.48×10⁻⁶

Chemical shift (δ) value  for 2nd proton=3.57PPM=3.57×10⁻⁶

We know that frequency of reference that is of TMS(Tetramethylsilane) is assumed to be 0.

We have to calculate the frequency of absorption of each protons downfield from TMS.

The formula for the chemical shift (δ) is:

δ=[Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

So using the above formula we can calculate the frequency of absorption for the two protons whose δ value is given.

1. For the proton having δ value 1.48ppm:

1.48= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

1.48×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]

1.48×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]

740Hz=[Frequency of sample(νₐ)]

2. For the proton having δ value 3.57ppm:

3.57= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

3.57×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]

3.57×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]

1785Hz=[Frequency of sample(νₐ)]

So the frequency of absorption for the proton having  δ value 1.48ppm is 740 Hz and for the proton having  δ value 3.57 ppm is 1785Hz.

Final answer:

To calculate the NMR frequencies for chloroethane at 1.48 ppm and 3.57 ppm on a 500 MHz spectrometer, multiply each chemical shift by 500 MHz. The results are 740 Hz for 1.48 ppm and 1785 Hz for 3.57 ppm.

Explanation:

The question pertains to calculating the frequency of each absorption in the 1H NMR spectrum of chloroethane, observed at chemical shifts of 1.48 ppm and 3.57 ppm, on a 500 MHz NMR spectrometer. The frequency of an absorption in Hz can be calculated by multiplying the parts per million (ppm) chemical shift by the frequency of the spectrometer in MHz. To calculate the frequency downfield from TMS (tetramethylsilane), the chemical shifts in ppm are multiplied by the spectrometer frequency.

For the 1.48 ppm peak
500 MHz Spectrometer

= 1.48 ppm *500 MHz

= 740 Hz

For the 3.57 ppm peak:

= 3.57 ppm *500 MHz

= 1785 Hz

Therefore, the frequencies of the absorptions downfield from TMS for chloroethane on a 500 MHz NMR spectrometer are 740 Hz and 1785 Hz for the chemical shifts of 1.48 ppm and 3.57 ppm, respectively.

The answer is in the photo

The molarity of the luminol stock solution is calculated by dividing the number of moles of luminol (0.08464 moles) by the volume of the solution in liters (0.075 L), resulting in a molarity of 1.13 M.

The ideal way to characterise stock solutions is as concentrated solutions with precise, known concentrations that will be diluted for later usage in laboratories. Although you have the option to forego preparing stock solutions, doing so can simplify your workflow and save you a significant amount of time and money. To solve the question we need to first determine the number of moles of luminol and then use the volume of the solution to find the molar concentration.

The molecular weight of luminol (C₈H₇N₃O₂) is approximately 177.16 g/mol. The number of moles of luminol can be calculated by dividing the mass of luminol by its molecular weight:

Number of moles of luminol = Mass of luminol / Molecular weight of luminol
Number of moles of luminol = 15.0 g / 177.16 g/mol = 0.08464 moles

Next, we convert the volume of water from milliliters to liters:

Volume in liters = 75.0 mL × (1 L / 1000 mL) = 0.075 L

Now we can calculate the molarity of the luminol solution:

Molarity (M) = Number of moles of solute / Volume of solution in liters
Molarity (M) = 0.08464 moles / 0.075 L

Molarity (M) = 1.13 M

The stock solution of luminol has a molarity of 1.13 moles per liter (1.13 M).

Answer: [tex]H_{3}N-H-O-H[/tex] shows a hydrogen bond is the correct answer.

Explanation:

A hydrogen bond is defined as a weak bond that is formed between an electropositive atom (generally hydrogen atom) and an electronegative atom like oxygen, nitrogen and fluorine.

This bond is formed due to difference in the electronegativity of atoms present in a compound which contains a hydrogen bond. So, there occurs a partial positive charge on hydrogen atom and a partial negative charge on the electronegative atom.

For example, [tex]H_{3}N-H-O-H[/tex] molecule will have hydrogen bonds.

This is because both nitrogen and oxygen atoms are electronegative in nature. Therefore, partial charges will develop on the both electronegative and electropositive atoms of this compound.

Thus, we can conclude that out of the given options [tex]H_{3}N-H-O-H[/tex] shows a hydrogen bond.

A hydrogen bond is an intermolecular attractive force in which a hydrogen atom is attracted to a lone pair of electrons on an atom in a neighboring molecule. The correct choice that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H.

A hydrogen bond is an intermolecular attractive force in which a hydrogen atom, that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule. Hydrogen bonds are very strong compared to other dipole-dipole interactions, but still much weaker than a covalent bond. A typical hydrogen bond is about 5% as strong as a covalent bond.

Looking at the choices, H−H and H2O⋅⋅⋅⋅⋅⋅H−CH3 do not involve hydrogen bonding, as they lack a small, highly electronegative atom with a lone pair of electrons. H4C⋅⋅⋅⋅⋅⋅H−F does not involve hydrogen bonding either, as the fluorine atom is not highly electronegative enough to form a hydrogen bond. Therefore, the correct choice that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H.

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Answer:Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

Explanation:The mechanism of oxidation using Potassium permanganate

involves two steps   :

In the first step permanganate ion abstracts a alpha-hydrogen as hydride ion available at alcohol.

Alpha hydrogen is the hydrogen attached to the carbon bearing functional group.

In the second step the permanganate ion subsequently is reduced from +7 oxidation state of Manganese to +5 oxidation state of Manganese.

So for the oxidation of alcohol using potassium permanganate , there must be availability of alpha hydrogens.

we can also relate the order of oxidation with reference to number of alpha hydrogens present on the substrate.

so

Methanol has 3 available alpha hydrogen

Ethanol has 2 available alpha hydrogen  

2-propanol has 1 available alpha hydrogen as it is a secondary alcohol

2-methyl-2propanol has 0 available alpha hydrogens as it is a tertiary alcohol.

Greater the number of available hydrogens easier would be oxidation of alcohols so the order of oxidation would be the following:

Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

please refer the attachment for the structures of following compounds.

Answer: The amount of heat absorbed by aluminium is 2.271 kJ.

Explanation:

To calculate the amount of heat absorbed or released, we use the equation:

[tex]Q= m\times c\times \Delta T[/tex]

Q = heat absorbed  = ?

m = mass of aluminium = 55.5 g

c = specific heat capacity of aluminium = 0.930 J/g ° C      

Putting values in above equation, we get:

[tex]\Delta T={\text{Change in temperature}}=(67-23)^oC=44^oC[/tex]  

[tex]Q=55.5g\times 0.930J/g^oC\times 44^oC[/tex]

Q = 2271.06 Joules

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 2271.06 J will be equal to 2.271 kJ

Hence, the amount of heat absorbed by aluminium is 2.271 kJ.

Answer: 0.07 moles

Explanation:

According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.

Thus [tex]p_{total}=p_{H_2O}+p_{H_2}[/tex]

Given: [tex]p_{total}=725torr[/tex]

[tex]p_{H_2O}=24torr[/tex]

[tex]p_{H_2}=?[/tex]

Thus [tex]725torr=24torr+p_{H_2}[/tex]

[tex]p_{H_2}=701torr[/tex]

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P= Pressure of the gas = 701 torr = 0.92 atm     (760torr=1atm)

V= Volume of the gas = 1.85 L

T= Temperature of the gas = 308 K

R= Value of gas constant = 0.082 Latm\K mol

[tex]n=\frac{PV}{RT}=\frac{0.92\times 1.85L}{0.0821 \times 308}=0.07moles[/tex]

Thus 0.07 moles of hydrogen gas is contained in 1.85 L of this mixture at 308 K.

Answer:

0.00191 meter is the diameter of the sphere.

Explanation:

Velocity of the rigid sphere = [tex]v=\frac{1m}{500 s}[/tex]

Density of the fluid = [tex]\rho _f=1000 kg/m^3[/tex]

Density of the particle= [tex]\rho _p=1100 kg/m^3[/tex]

Viscosity of the fluid =[tex]\eta =100 centiPoise = 1 Poise=0.1 Kg/(m s)[/tex]

Radius of the sphere  = r

[tex]v=\frac{2}{9}\times \frac{\rho _p-\rho _f}{\eta }gr^2[/tex]

(Terminal velocity)

[tex]\frac{1 m}{500 s}=\frac{2}{9}\times \frac{1100 kg/m^3-1000 kg/m^3}{0.1 kg/(m s)}\times 9.8 m/s^2\times r^2[/tex]

r = 0.00191 m

0.00191 meter is the diameter of the sphere.

Answer: A. PH=4.50

Explanation:

As we know that pH=-log(H3O+)

pH=-log(3.2x10^-5)

PH= -log(3.2)+log(10^-5)

=-(0.50+(-5))

PH= -(0.5-5)=-(-4.5)

pH=4.50

B. If we know pH , then [H3O+]= 10^-pH

To find the pH of a substance, use pH = -log[H3O+]. Given [H3O+] = 3.2 × 10-5, the pH ≈ 4.49. To find hydrogen ion concentration from pH, use [H3O+] = 10^-pH.

pH is a measure of the acidity or alkalinity of a solution, defined as the negative logarithm of the hydrogen ion concentration (expressed in molarity) of a solution.

To find the pH of a substance, you can use the formula: pH = -log[H3O+].

a. Given [H3O+] = 3.2 × 10-5, the pH is calculated as follows:

b. To find the hydrogen ion concentration given a pH value, you rearrange the formula as: [H3O+] = 10-pH.

For example, if the pH of a solution is 3:

Therefore, solutions with a pH of less than seven are acidic, while those with a pH greater than seven are basic (alkaline). The pH scale ranges from 0 to 14 and a pH of 7 is considered neutral.

Answer:

[BrCl]=0.02934M

[Br2]=[Cl2]=0.03533M

Explanation:

First, consider the Kc definition:

[tex]1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}[/tex]

It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;

[tex]n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x\\[/tex]

Where the zero superscript means the initial moles.

Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:

[tex]M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x\\[/tex]

And now 'x' is a change in molarity.

Replacing these in the Kc equation we have:

[tex]1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}[/tex]

Where the only unknown is 'x'. So, let's solve the equation:

[tex]1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533[/tex]

The result [tex]x_{1}=0.0555[/tex] lacks of sense because it will give a negative concentration for BrCl, so the result is [tex]x_{2}=0.00533[/tex].

Applying the result, the concentrations at equilibrium are:

[tex]M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M[/tex]

If you calculate Kc with this concentrations it will give 1.45 as a result.

Greets, I will be happy to solve any doubt you have.

To calculate the equilibrium concentrations for a given reaction with initial concentrations provided, use the equilibrium constant Kc to determine the equilibrium concentrations of the reactants and products.

Equilibrium concentrations calculation:

Given initial concentrations: [BrCl] = 0.0400 M, [Br2] = 0.0300 M, and [Cl2] = 0.0300 M, use the equilibrium constant Kc = 1.45 to calculate the equilibrium concentrations of Br2, Cl2, and BrCl in the reaction equation BrCl(g) = Br2(g) + Cl2(g). Work out the concentrations using the stoichiometry of the reaction.

Explanation:

Meso compounds are optically inactive stereoisomers. Despite of having chiral carbon, they do not show  optical activity because it has a plane of symmetry in its structure itself. It is superposable on its mirror image.

Aldo sugars on reaction which reducing agents such as , NaBH₄ reduces the carbonyl group in the sugar to alcohol and gives corresponding alditol.

The D- aldohexose which on reduction gives a meso alditol are allose and galactose.

The structure is shown in the image below. Thus, in allitol and galactitol formed have internal plane of symmetry which makes the optically inactive.

Answer:

Explanation:

The compound is an alkene (cycloalkene), with a methyl group as substituent (it substitutes one hydrogen in the carbon chain).

The IUPAC's rules state that the location of the carbon-carbon double bond in the structure is indicated by specifying the number of the carbon atom at which the C=C bond starts, assigning the lowest possible number to the double bond: this in this case is number 1.

Cyclohexene is the main chain and mehtyl is a substituent, as already said.

So, the name 1-methyl-3-cychlohexene is not valid (position 1 is for the carbon-carbon double bond).

The name methycyclohexene is not valid because it is not telling the position of methylgroup.

The name 5-methylcyclohexene is not valid because the position five should be named 2 in the cyclohexene (you must use the smallest number), so the name should be 2-methyl... instead of 5 methyl...

1-methyl-4cyclohexene is not valid because, as said, the position 1 is reserved for the carbon-carbon double bond.

Only 4-methylcyclohexene is a valid name.

The file attached shows the structure. I have added numbers on the carbons of the main chain to show you how that the methyl group is in the positiion number 4.

What is the rate law for the uncatalyzed reaction?

In this type of reactions the rate law is dependent on the concentration of the reactants. In this case Ce and Tl ions. The answer will be rate=k[Ce[tex]^{4+}[/tex]][tex]^{2}[/tex][Tl[tex]^{+}[/tex]] because according to the rate law the reactants concentration should be powered at the number of moles, and from the equation we deduce that Ce[tex]^{4+}[/tex] concentration will be powered to 2.

the other answers are wrong  

If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction?

“All reactions that occur in one step are slow” - it is not correct because there are one step reactions that are fast.

“The reaction requires the collision of three particles with the correct energy and orientation” - it is the correct answer because the particles should collide in exactly the same time with the required amount of energy for the reaction to take place. The probability of an effective three-particle collision is low so the reaction will be a slow one.

“The transition state is low in energy” -  in this case the rate of the reaction should be fast because the energy for the reaction needed to take place is low.

The catalyzed reaction is first order in [Ce[tex]^{4+}[/tex]] and first order in [Mn[tex]^{2+}[/tex]]. Which of the steps in the catalyzed mechanism is rate determining?

rate =k[Ce[tex]^{4+}[/tex]] [Mn[tex]^{2+}[/tex]] because the rate of the reaction may be done by increasing the amount of the catalyst (Mn[tex]^{2+}[/tex]).

The catalyzed reaction is first order in which of the steps in the catalyzed mechanism is rate determining?

Ce[tex]^{4+}[/tex][tex]_{(aq)}[/tex]+Mn[tex]^{2+}[/tex][tex]_{(aq)}[/tex]→Ce[tex]^{3+}[/tex][tex]_{(aq)}[/tex]+Mn[tex]^{3+}[/tex][tex]_{(aq)}[/tex]

Answer:

[tex]\boxed{\text{c. The concentration of reactants is much greater than that of products.}}[/tex]

Explanation:

NO₂ + NO₃ ⇌ N₂O₅; K = 2.1 × 10⁻²⁰

We often write K as

[tex]K = \dfrac{[\text{Products}]}{[\text{Reactants}]}[/tex]

If K is large, more of the molecules exist as products.

If K is small, more of the molecules exist as reactants.

[tex]\text{Since K is small, }\\\boxed{\textbf{the concentration of reactants is much greater than that of products.}}[/tex]

a. b., and d. are wrong. The concentration of reactants is greater.

Final answer:

The equilibrium constant of K = 2.1 × 10−20 suggests that at equilibrium, the reactants' concentration significantly surpasses the products', confirming option (c) as correct.

Explanation:

The equilibrium constant for the reaction NO2(g) + NO3(g) ⇌ N2O5(g) is K = 2.1 × 10−20. An equilibrium constant of such a small magnitude indicates that, at equilibrium, the concentration of reactants (NO2 and NO3) will be much greater than the concentration of products (N2O5). This means that option (c) is correct: At equilibrium the concentration of reactants is much greater than that of products. It does not mean reactions stop; both the forward and reverse reactions continue occurring at equal rates, maintaining the equilibrium concentrations of reactants and products.

Answer : The correct option is, Both gases have the same average kinetic energy.

Explanation :

As we are given that there are 8 moles of 'He' and 'Xe'.

At STP condition,

The temperature and pressure are 273 K and 1 atm respectively.

The formula of average kinetic energy is, [tex]K.E=\frac{3}{2}RT[/tex]. From this we conclude that the average kinetic energy is depends on the temperature only. So, at same temperature the average kinetic energy will also be same for both the gases.

As we know that the molecular speed is inversely proportional to the square root of the molecular mass. That means, it depends on the molar mass of substance. So, both the gases have the different molecular speed.

As we know that the density is directly proportional to the mass of substance. That means, it depends on the mass of substance. So, both the gases have the different density.

At STP,

As, 1 mole of He gas contains 22.4 liter volume of He gas

So, 8 mole of He gas contains [tex]8\times 22.4=179.2[/tex] liter volume of He gas

As, 1 mole of Xe  gas contains 22.4 liter volume of Xe gas

So, 8 mole of Xe gas contains [tex]8\times 22.4=179.2[/tex] liter volume of Xe gas

The mixture of has volume = 179.2 + 179.2 = 358.4 L

Hence, from the above we conclude that the correct option is, Both gases have the same average kinetic energy.